A wave front AB moving in air is incident on a plane glass surface. XY , as given in the figure, its position CD after refraction through glass slab is shown along with normal drawn at A and D. the refractive index of glass w.r.t air is given by

Options

A. `(sin theta)/(sin theta)`
B. `(sin theta)/(sin phi)`
C. `(sin phi)/(sin theta)`
D. `(AB)/(CD)`

Answer

To find the refractive index of glass with respect to air, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Angles**:
- Let \( \theta \) be the angle of incidence (the angle between the incident wave front AB and the normal at point A).
- Let \( \theta' \) be the angle of refraction (the angle between the refracted wave front CD and the normal at point D).

2. **Use Snell's Law**:
- According to Snell's Law, the relationship between the angles of incidence and refraction is given by:
\[ \mu = \frac{\sin \theta}{\sin \theta'} \]
where \( \mu \) is the refractive index of the medium (glass) with respect to the other medium (air).

3. **Determine the Refractive Index**:
- The refractive index of glass with respect to air can be expressed as:
\[ \mu = \frac{V_a}{V_g} \]
where \( V_a \) is the speed of light in air and \( V_g \) is the speed of light in glass.

4. **Relate the Distances in the Triangles**:
- In triangle ABD, the length BD can be expressed as:
\[ BD = AD \cdot \sin \theta \]
- In triangle ACD, the length AC can be expressed as:
\[ AC = AD \cdot \sin \theta' \]

5. **Substitute the Lengths into the Refractive Index Formula**:
- From the triangles, we can write:
\[ \frac{BD}{AC} = \frac{AD \cdot \sin \theta}{AD \cdot \sin \theta'} = \frac{\sin \theta}{\sin \theta'} \]
- Thus, we can conclude that:
\[ \mu = \frac{\sin \theta}{\sin \theta'} \]

6. **Final Answer**:
- Therefore, the refractive index of glass with respect to air is:
\[ \mu = \frac{\sin \theta}{\sin \theta'} \]